Prove the following identity, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A} = \dfrac{2}{1 - 2\text{cos }^2A} = \dfrac{2\text{ sec }^2A}{\text{tan }^2A - 1}$

Given,

$\Rightarrow \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}}\\[1em] \Rightarrow \dfrac{\text{(sin A + cos A)}^2 + \text{(sin A - cos A)}^2}{(\text{sin} A - \text{cos} A)(\text{sin A + cos A})}\\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + 2\text{sin A.cos A} + \text{sin}^2 A + \text{cos}^2 A - 2\text{sin A.cos A}}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2\text{sin}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2(1 - \text{cos}^2 A) + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2 - 2\text{cos}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}$

So proved,

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A}$

Now,

$\Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{(1 - \text{cos}^2 A) - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{1 - \text{cos}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{1 - 2\text{cos}^2 A}$

So proved,

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{1 - 2\text{cos}^2 A}$

Now,

$\Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}$

Dividing numerator and denominator by cos² A, we get :

$\Rightarrow \dfrac{\dfrac{2}{\text{cos}^2 A}}{\dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \dfrac{\text{cos}^2 A}{\text{cos}^2 A}}\\[1em] \Rightarrow \dfrac{2\text{sec}^2 A}{\text{tan}^2 A - 1}\\[1em]$

So proved,

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2\text{sec }^2A}{\text{tan }^2A - 1}$

Hence, proved that

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A} = \dfrac{2}{1 - 2\text{cos }^2A} = \dfrac{2\text{sec }^2A}{\text{tan }^2A - 1}$.